This question tests the skill of using average bond enthalpies to estimate the enthalpy change for a chemical reaction. To estimate ΔH, calculate the total energy required to break the bonds in the reactants and subtract the total energy released when forming the bonds in the products. For the reaction H2 + Cl2 → 2HCl, the bonds broken are one H-H bond (436 kJ/mol) and one Cl-Cl bond (243 kJ/mol), totaling 679 kJ/mol. The bonds formed are two H-Cl bonds (2 × 431 kJ/mol = 862 kJ/mol), so ΔH = 679 - 862 = -183 kJ/mol, indicating an exothermic reaction. A tempting distractor is +183 kJ/mol, which results from the misconception of subtracting the energy of bonds broken from the energy of bonds formed instead of the reverse. A transferable strategy is to always remember that ΔH ≈ Σ(bond enthalpies broken) - Σ(bond enthalpies formed), where a negative value means heat is released.

This question tests the skill of estimating the enthalpy change of a reaction using average bond enthalpies. For the combustion of methane, the bonds broken are four C–H bonds (413 kJ/mol each) and two O=O bonds (498 kJ/mol each), totaling 2648 kJ/mol. The bonds formed are two C=O bonds (799 kJ/mol each) and four O–H bonds (463 kJ/mol each), totaling 3450 kJ/mol. Thus, ΔH ≈ 2648 - 3450 = -802 kJ/mol. A tempting distractor is +802 kJ/mol, arising from the misconception of subtracting in the wrong order. A transferable strategy is to list all bonds in reactants and products systematically to ensure none are missed in complex molecules.

This question tests the skill of using average bond enthalpies to estimate the enthalpy change for a chemical reaction. To estimate ΔH, calculate the total energy required to break the bonds in the reactants and subtract the total energy released when forming the bonds in the products. For the reaction H2 + F2 → 2HF, the bonds broken are one H-H bond (436 kJ/mol) and one F-F bond (159 kJ/mol), totaling 595 kJ/mol. The bonds formed are two H-F bonds (2 × 565 kJ/mol = 1130 kJ/mol), so ΔH = 595 - 1130 = -535 kJ/mol, indicating a strongly exothermic reaction. A tempting distractor is +535 kJ/mol, which results from the misconception of reversing the subtraction, calculating bonds formed minus bonds broken. A transferable strategy is to double-check the arithmetic and ensure all bonds are accounted for according to the balanced equation.

This question tests the skill of using average bond enthalpies to estimate the enthalpy change (ΔH) for a chemical reaction. The reaction is H₂(g) + Cl₂(g) → 2HCl(g), so bonds broken are one H-H bond at 436 kJ/mol and one Cl-Cl bond at 243 kJ/mol, totaling 679 kJ/mol of energy absorbed. Bonds formed are two H-Cl bonds at 431 kJ/mol each, totaling 862 kJ/mol of energy released. Therefore, the estimated ΔH is 679 - 862 = -183 kJ/mol, indicating an exothermic process as the energy released exceeds the energy absorbed. A tempting distractor is +183 kJ/mol, which stems from the misconception of reversing the subtraction order, treating formed bonds as energy absorbed instead of released. A transferable strategy is to always list bonds broken (positive) and formed (negative) based on the balanced equation and apply the formula ΔH ≈ Σ(bonds broken) - Σ(bonds formed) while double-checking multiples for accurate estimation.

This question tests the skill of using average bond enthalpies to estimate the enthalpy change (ΔH) for a chemical reaction. The reaction is CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g), so bonds broken are one C-H bond at 413 kJ/mol and one Cl-Cl bond at 243 kJ/mol, totaling 656 kJ/mol of energy absorbed. Bonds formed are one C-Cl bond at 328 kJ/mol and one H-Cl bond at 431 kJ/mol, totaling 759 kJ/mol of energy released. Therefore, the estimated ΔH is 656 - 759 = -103 kJ/mol, indicating an exothermic process as the energy released exceeds the energy absorbed. A tempting distractor is +103 kJ/mol, which stems from the misconception of reversing the subtraction order, treating formed bonds as energy absorbed instead of released. A transferable strategy is to always list bonds broken (positive) and formed (negative) based on the balanced equation and apply the formula ΔH ≈ Σ(bonds broken) - Σ(bonds formed) while double-checking multiples for accurate estimation.

This question tests the skill of using average bond enthalpies to estimate the enthalpy change (ΔH) for a chemical reaction. The reaction is H₂(g) + Br₂(g) → 2HBr(g), so bonds broken are one H-H bond at 436 kJ/mol and one Br-Br bond at 193 kJ/mol, totaling 629 kJ/mol of energy absorbed. Bonds formed are two H-Br bonds at 366 kJ/mol each, totaling 732 kJ/mol of energy released. Therefore, the estimated ΔH is 629 - 732 = -103 kJ/mol, indicating an exothermic process as the energy released exceeds the energy absorbed. A tempting distractor is +103 kJ/mol, which stems from the misconception of reversing the subtraction order, treating formed bonds as energy absorbed instead of released. A transferable strategy is to always list bonds broken (positive) and formed (negative) based on the balanced equation and apply the formula ΔH ≈ Σ(bonds broken) - Σ(bonds formed) while double-checking multiples for accurate estimation.

This question tests the skill of estimating the enthalpy change of a reaction using average bond enthalpies. For this addition reaction, the bonds broken are one C=C bond (614 kJ/mol) and one Cl–Cl bond (243 kJ/mol), summing to 857 kJ/mol. The bonds formed are one C–C bond (347 kJ/mol) and two C–Cl bonds (338 kJ/mol each), summing to 1023 kJ/mol. Thus, ΔH ≈ 857 - 1023 = -166 kJ/mol. A tempting distractor is +166 kJ/mol, stemming from the misconception of reversing the formula's subtraction. A transferable strategy is to confirm that the bond counts match the molecular changes in addition reactions to multiple bonds.

This question tests the ability to estimate the enthalpy change (ΔH) of a reaction using average bond enthalpies. For the reaction H₂(g) + Br₂(g) → 2HBr(g), the bonds broken are one H–H bond (436 kJ/mol) and one Br–Br bond (193 kJ/mol), totaling 629 kJ/mol. The bonds formed are two H–Br bonds (366 kJ/mol each), totaling 732 kJ/mol. Therefore, ΔH ≈ 629 - 732 = -103 kJ/mol, showing the reaction is exothermic due to stronger bonds formed compared to those broken. A tempting distractor is +103 kJ/mol, stemming from the misconception of subtracting broken from formed enthalpies, leading to an incorrect positive sign. A transferable strategy is to account for the stoichiometry by multiplying bond enthalpies by the number of bonds, then compute ΔH as sum(broken) - sum(formed) for reliable estimates in gaseous reactions.

This question tests the skill of estimating the enthalpy change of a reaction using average bond enthalpies. To estimate ΔH, calculate the total energy required to break the bonds in the reactants and subtract the total energy released when forming the bonds in the products. For this reaction, the bonds broken are one H–H bond (436 kJ/mol) and one Cl–Cl bond (243 kJ/mol), summing to 679 kJ/mol. The bonds formed are two H–Cl bonds (431 kJ/mol each), summing to 862 kJ/mol, so ΔH ≈ 679 - 862 = -183 kJ/mol. A tempting distractor is +183 kJ/mol, which results from the misconception of reversing the subtraction, treating bond formation as energy input instead of release. A transferable strategy is to consistently remember that bond breaking is endothermic and bond forming is exothermic, ensuring the correct sign for ΔH.

This question tests the skill of using average bond enthalpies to estimate the enthalpy change (ΔH) for a chemical reaction. The reaction is CH₂=CH₂(g) + HCl(g) → CH₃CH₂Cl(g), so bonds broken are one C=C bond at 614 kJ/mol and one H-Cl bond at 431 kJ/mol, totaling 1045 kJ/mol of energy absorbed. Bonds formed are one C-C bond at 348 kJ/mol, one C-H bond at 413 kJ/mol, and one C-Cl bond at 328 kJ/mol, totaling 1089 kJ/mol of energy released. Therefore, the estimated ΔH is 1045 - 1089 = -44 kJ/mol, indicating an exothermic process as the energy released exceeds the energy absorbed. A tempting distractor is +44 kJ/mol, which stems from the misconception of reversing the subtraction order, treating formed bonds as energy absorbed instead of released. A transferable strategy is to always list bonds broken (positive) and formed (negative) based on the balanced equation and apply the formula ΔH ≈ Σ(bonds broken) - Σ(bonds formed) while double-checking multiples for accurate estimation.